# LeetCode 339Nested List Weight Sum

## LeetCode 339 Nested List Weight Sum Problem

``````# """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
# class NestedInteger(object):
#    def isInteger(self):
#        """
#        @return True if this NestedInteger holds a single integer, rather than a nested list.
#        :rtype bool
#        """
#
#    def getInteger(self):
#        """
#        @return the single integer that this NestedInteger holds, if it holds a single integer
#        Return None if this NestedInteger holds a nested list
#        :rtype int
#        """
#
#    def getList(self):
#        """
#        @return the nested list that this NestedInteger holds, if it holds a nested list
#        Return None if this NestedInteger holds a single integer
#        :rtype List[NestedInteger]
#        """

class Solution(object):
def depthSum(self, nestedList):
"""
:type nestedList: List[NestedInteger]
:rtype: int
"""
return self.depthSum_helper(nestedList, 1)

def depthSum_helper(self, nestedList, depth):
res = 0
for l in nestedList:
if l.isInteger():
res += l.getInteger() * depth
else:
res += self.depthSum_helper(l.getList(), depth + 1)
return res
``````

## List of all Nested List Weight Sum problems

Leetcode 339 Nested List Weight Sum problem solution in python3 with explanation. This is the best place to expand your knowledge and get prepared for your next interview.

"""

class NestedInteger(object):

def isInteger(self):

"""

@return True if this NestedInteger holds a single integer, rather than a nested list.

:rtype bool

"""

def getInteger(self):

"""

@return the single integer that this NestedInteger holds, if it holds a single integer

Return None if this NestedInteger holds a nested list

:rtype int

"""

def getList(self):

"""

@return the nested list that this NestedInteger holds, if it holds a nested list

Return None if this NestedInteger holds a single integer

Feedback is the most important part of any website.